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A particle P of weight W newtons is attached to one end of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 1 - 2014 - Paper 1

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A particle P of weight W newtons is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point O. A horizontal forc... show full transcript

Worked Solution & Example Answer:A particle P of weight W newtons is attached to one end of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 1 - 2014 - Paper 1

Step 1

Find (a) the tension in the string.

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Answer

To find the tension TT in the string, we resolve the horizontal forces acting on the particle P.

Given that the horizontal force is 5 N, we have:

5=Tcos(25)5 = T \cos(25^\circ)

From this equation, we can solve for TT:

T=5cos(25)T = \frac{5}{\cos(25^\circ)}

Now, calculating this value:

T5/0.90635.51 NT \approx 5 / 0.9063 \approx 5.51 \text{ N}

Thus, the tension in the string is approximately 5.51 N.

Step 2

Find (b) the value of W.

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Answer

Next, we resolve the vertical forces acting on particle P. The weight WW acts downwards while the vertical component of the tension acts upwards. We can set the equilibrium condition:

W=Tsin(25)W = T \sin(25^\circ)

Replacing TT from the previous step, we have:

W=(5cos(25))sin(25)W = \left(\frac{5}{\cos(25^\circ)}\right) \sin(25^\circ)

Evaluating:

W=5sin(25)cos(25)50.46631.18850.55472.77 NW = 5 \cdot \frac{\sin(25^\circ)}{\cos(25^\circ)} \approx 5 \cdot 0.4663 \cdot 1.188 \approx 5 \cdot 0.5547 \approx 2.77 \text{ N}

Finally, the value of WW is approximately 2.77 N.

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